Đáp án:

$1.\; V_{H_2(\min)}=2,24\text{(l)}\\ 2. \; a)\; 8\text{(g)}\qquad b)\; 2,24\text{(l)}\\ 3.\; Sai\; de$

Giải thích các bước giải:

1. \(n_{CuO}=\dfrac 8{80}=0,1\text{(mol)}\\ CuO+H_2\xrightarrow{t^{\circ}}Cu+H_2O\\ n_{H_2(\min)}=n_{CuO}=0,1\text{(mol)}\\\to V_{H_2(\min)}=0,1\times 22,4=2,24\text{(l)}\)

2.  \(n_{Mg}=\dfrac{4,8}{24}=0,2\text{(mol)}\\ 2Mg+O_2\xrightarrow{t^{\circ}}2MgO\\a)\;n_{MgO}=n_{Mg}=0,2\text{(mol)}\\\to m_{MgO}=40\times 0,2=8\text{(g)}\\ b)\;n_{O_2}=\dfrac 12\times n_{Mg}=\dfrac 12\times 0,2=0,1\text{(mol)}\\\to V_{O_2}=0,1\times 22,4=2,24\text{(l)}\)

3. \(\begin{cases}n_{CO_2}=1\text{(mol)}\to n_C=1\text{(mol)}; n_O=1\cdot 2=2\text{(mol)}\\n_{H_2O}=1\text{(mol)}\to n_H=1\cdot 2=2\text{(mol)}; n_O=1\text{(mol)}\end{cases}\)

Bảo toàn mol O: $n_{O(A)}+2n_{O_2}=n_{O(CO_2)}+n_{O(H_2O)}\\\to n_{O(A)}=2+1-2\times 2=-1\text{(mol)}$

$\to$Sai đề

1/

$n_{CuO}=8/80=0,1mol$

$PTHH :$

$CuO+H_2\overset{t^o}\to Cu+H_2O$

Theo pt :

$n_{Cu}=n_{H_2}=n_{CuO}=0,1mol$

$⇒m_{Cu}=0,1.64=6,4g$

$V_{H_2}=0,1.22,4=2,24l$

2/

$n_{Mg}=4,8/24=0,2mol$

$PTHH :$

$2Mg+O_2\overset{t^o}\to 2MgO$

$\text{a/Theo pt :}$

$n_{MgO}=n_{Mg}=0,2mol$

$⇒m_{MgO}=0,2.40=8g$

$\text{b/Theo pt :}$

$n_{O_2}=1/2.n_{Mg}=1/2.0,2=0,1mol$

$⇒V_{O_2}=0,1.22,4=2,24l$